Posted by **Anonymous** on Saturday, July 27, 2013 at 6:55am.

the dimensions of a rectangle are such that its length is 5 in more than its width. If the length were doubled and the width were decreases by 2 in, the area would be increased 136 in^2,what are the length and width of the rectangle

- algebra -
**MathMate**, Saturday, July 27, 2013 at 7:19am
Initially

w=width

w+5=length

w(w+5)=original area

Subsequently:

w-2=width

2(w+5)=length

(w-2)*2(w+5)=2*original area=w(w+5)=area

Therefore

2(w-2)(w+5)=w(w+5)

Solve for w:

2(w-2)=w (w cannot be equal to -5)

2w-4=w

w=4

- algebra -
**Reiny**, Saturday, July 27, 2013 at 8:27am
original width -- x

original length -- x+5

area = x(x+5)

new width = x-2

new length = 2x+10

new area = (2x+10)(x-2)

(2x+10)(x-2) - x(x+5) = 136

2x^2 + 6x - 20 - x^2 - 5x - 136 = 0

x^2 + x - 156 = 0

(x-12)(x+13) = 0

x = 12 or x = -13

ignoring the negative,

**width = 12
**

length = 17

check:

original area = 12(17) = 204

new width = 10

new length = 34

new area = 10(34) = 340

increase in area = 340-204 = 136 , YEahh!

- algebra -
**MathMate**, Saturday, July 27, 2013 at 4:23pm
Good catch!

Thank you!

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